Question

Find an equation of the tangent plane to the surface x y 2 + 3 x − z 2 = 4 at the point ( 2 , 1 , − 2 ) An equation of the tangent plane is

Answer #1

(a) Find an equation of the plane tangent to the surface xy ln x
− y^2 + z^2 + 5 = 0 at the point (1, −3, 2)
(b) Find the directional derivative of f(x, y, z) = xy ln x −
y^2 + z^2 + 5 at the point (1, −3, 2) in the direction of the
vector < 1, 0, −1 >. (Hint: Use the results of partial
derivatives from part(a))

Find an equation of the tangent plane to the surface given
parametrically by x = u^2, y = v^2, z = u+4v at the point (1, 4,
9).

8).
a) Find an equation of the tangent plane to the surface z = x at
(−4, 2, −1).
b) Explain why f(x, y) = x2ey is differentiable at (1, 0). Then
find the linearization L(x, y) of the function at that point.

Find the equation for the tangent plane to the surface
z=(xy)/(y+x) at the point P(1,1,1/2).

Find the equation of the tangent plane (in terms of x, y and z)
to the surface given by x = u, y = v and z = uv at the point (3, 2,
6).

Find an equation of the tangent plane to the given surface at
the specified point.
z = 2(x − 1)2 + 4(y + 3)2 +
1, (3, −1, 25)
Answer as z=

Find the equation of the tangent plane for the surface
represented by x = u2, y = u - v2, and z =
v2 at the following point (1,0,1).

Find an equation of the tangent plane to the given surface at
the specified point.
z = 2(x − 1)2 + 4(y + 3)2 +
9, (2, −2, 15)

Find an equation of the tangent plane to the surface z = x^2 +
xy + 3y^2 at the point (1, 1, 5)

Find the equation of the tangent plane to the surface determined
by x2y4+z−35=0 at x=3, y=4.

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